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\subsubsection*{Examples}

\begin{itemize}
    \item $\sum_{j = 1}^\infty \frac{j}{2^j}$.
        Then
        \[ \frac{a_{n + 1}}{a_n} = \frac{n +
        1}{2^{n + 1}} \cdot \frac{2^n}{n} =
        \frac{n + 1}{2n} \to \frac{1}{2} < 1 .\]
        So we have convergence by ratio test.
    \item $\sum_{j = 1}^\infty \frac{1}{n}$
        diverges, and $\sum_{j = 1}^\infty
        \frac{1}{n^2}$ converges. Note ratio test
        gives limit 1 in both cases, so
        \emph{inconclusive} if limit is 1. Since
        $n^{1 / n} \to 1$ as $n \to \infty$, the
        root test is also inconclusive when limit
        is 1. To see this limit, write
        \[ n^{1 / n} = 1 + \delta_n , \qquad
        \delta_n > 0 .\]
        \[ n = (1 + \delta_n)^n > \frac{n(n -
        1)}{2} \delta_n^2 \]
        (binomial expansion)
        \[ \implies \delta_n^2 < \frac{2}{n - 1}
        \implies \delta_n \to 0 \]
    \item $\sum_{j = 1}^\infty \left[ \frac{n +
        1}{3n + 5} \right]^n$ converges by
        root test since
        \[ \frac{n + 1}{3n + 5} \to \frac{1}{3} <
        1 .\]
\end{itemize}

\noindent
Another useful test:

\begin{theorem}[Cauchy's Condensation Test]
    Let $a_n$ be a decreasing sequence of positive
    terms. Then $\sum_{j = 1}^\infty a_n$
    converges if and only if
    \[ \sum_{j = 1}^\infty 2^n a_{2^n} \]
    converges.
\end{theorem}

\begin{proof}
    First we observe that if $a_n$ is decreasing,
    then
    \[ a_{2^k} \le a-{2^{k - 1} + i} \le a_{2^{k -
    1}}, \qquad 1 \le i \le 2^{k - 1} \]
    (for any $k \ge 1$.) Assume now that $\sum_{j
    = 1}^\infty a_j$ converges with sum $A$. Then
    \[ 2^{n - 1} a_{2^n} \le a_{2^{n - 1} + 1} +
    a_{2^{n - 1} + 2} + \cdots + a_{2^n} = \sum_{m
    = 2^{n - 1}}^{2^n} a_m .\]
    Thus
    \[ \sum_{n = 1}^N 2^{n - 1} a_{2^n} \le
    \sum_{n = 1}^N \sum_{m = 2^{n - 1} + 1}^{2^n}
    a_m = \sum_{m = 2}^{2^N} a_m .\]
    \[ \implies \sum_{n = 1}^N 2^n a_{2^n} \le 2
    \sum_{m = 2}^{2^N} a_m \le 2(A - a_1) \]
    Thus $\sum_{n = 1}^N 2^n a_{2^n}$ being
    increasing and bounded above,
    \emph{converges}. Conversely, assume that
    $\sum_{j = 1}^\infty 2^j a_{2^j}$ converges.
    Then
    \[ \sum_{m = 2^{n - 1} + 1}^{2^n} a_m \le
    \sum_{m = 2^{n - 1} + 1}^{2^n} a_{2^{n - 1}} =
    2^{n - 1} a_{2^{n - 1}} .\]
    \[ \implies \sum_{m = 2}^{2^N} a_m = \sum_{n =
    1}^N \sum_{m = 2^{n - 1} + 1} a_m \le \sum_{n
    = 1}^N 2^{n - 1} a_{2^{n - 1}} \le B .\]
    So $\sum_{m = 1}^N a_m$ is a bounded
    increasing sequence and thus it converges.
\end{proof}

\subsubsection*{Examples / Applications}

\begin{claim*}
    $\sum_{j = 1}^\infty \frac{1}{n^k}$ converges
    if and only if $k > 1$.
\end{claim*}

\begin{proof}
    Note that it is a decreasing sequence of
    positive terms.
    \[ \frac{1}{(n + 1)^k} < \frac{1}{n^k}, \qquad
    \left( \frac{n}{n + 1} \right)^k < 1 \]
    Now:
    \[ 2^n a_{2^n} = 2^n \left[ \frac{1}{2^n}
    \right]^k = 2^{n - nk} = (2^{1 - k})^n \]
    so it is a geometric series with ratio $2^{1 -
    k}$, and it converges if and only if $2^{1 -
    k} < 1$, so if and only if $k > 1$.
\end{proof}

\subsubsection*{Alternating Series}

\begin{theorem}[Alternating Series Test]
    If $a_n$ decreases and tends to zero as $n \to
    \infty$, then the series $\sum_{j = 1}^\infty
    (-1)^{n + 1} a_n$ converges.
\end{theorem}

\begin{example*}
    $\sum_{n = 1}^\infty \frac{(-1)^{n + 1}}{n}$
    converges.
\end{example*}

\begin{proof}
    Let $s_n = a_1 - a_2 + \cdots + (-1)^{n + 1}
    a_n$. Note
    \[ s_{2n} = (a_1 - a_2) + (a_3 - a_4) + \cdots
    + \ub{(a_{2n - 1} - a_{2n})}_{\ge 0} \ge s_{2n
    - 2} \]
    \[ s_{2n} = a_1 - (a_2 - a_3) - (a_4 - a_5) -
    \cdots - (a_{2n - 2} - a_{2n - 1}) - a_{2n}
    \le a_1 \]
    So $s_{2n}$ is increasing and bounded above,
    so $s_{2n} \to s$. Also note $s_{2n + 1} =
    s_{2n} + a_{2n + 1} \to s + 0 = s$. This
    implies that $s_n$ converges to $s$: \\
    Given $\eps > 0$, there exists $N_1$ such that
    for all $n \ge N_1$, $|s_{2n} - s| < \eps$ and
    there exists $N_2$ such that for all $n \ge
    N_2$, $|s_{2n + 1} - s| < \eps$. Take $N =
    2\max\{N_1, N_2\} + 1$. Then if $k \ge N$, we
    have $|s_k - s| < \eps$, so $s_k \to s$.
\end{proof}